$f(x, y) = \dfrac{e^{2y}}{2x}$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{e^{2y}(2x - 1)}{2x^2}$ (Choice B) B $\dfrac{-e^{2y}}{x^2}$ (Choice C) C $\dfrac{e^{2y}}{x}$ (Choice D) D $\dfrac{-e^{2y}}{2x^2}$
Solution: Taking a partial derivative with respect to $y$ means treating $x$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial y} &= \dfrac{\partial}{\partial y} \left[ \dfrac{e^{{2y}}}{2x} \right] \\ \\ &= \dfrac{1}{2x} (2e^{{2y}}) \\ \\ &= \dfrac{e^{{2y}}}{x} \end{aligned}$ In conclusion, $\dfrac{\partial f}{\partial y} = \dfrac{e^{2y}}{x}$